What is the area of the parallelogram?

Let us check the transformations. The first step is regarding the argument of the function, the -2x part. So, first of all, the minus sign implies that the function is reflected along the y-axis since f(x) is replaced with f(-x). However, cosx is symmetric along that axis so there is no change on the graph. Also, the 2 factor means that the function is compressed along the x-axis, since now f(2) corresponds to f(1) etc. (if we substitute 1 in the cos(2x), it is as if substituting 2 in the origninal function cosx). Finally, we have that the factor 3 in front of cos, implies that the function is dilated along the y-axis; the highs become 3 times higher and the lows 3 times low.

4 0

3 years ago

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A jar made of 3/16-inch-thick glass has an inside radius of 3.00 inches and a total height of 6.00 inches (including the bottom

steposvetlana [31]

We can calculate the volume of a glass shell of the jar by calculating the total volume of a jar and then subtract the volume of air that fits in the jar (the space bounded by the shell).
The total volume of the jar is:
$V_t=Bh=\pi h(r+a)^2$
Where h is the height of the jar, a is the thickness and r is the inner radius.
The volume of empty space, bounded by the jar, is:
$V_e=B(h-a)=\pi r^2(h-a)$
The volume of a jar shell is the difference between these two:
$V_s=V_t-V_e=\pi h(r+a)^2-\pi r^2(h-a)$
When we plug in all the number we get:
$V_s=27.17 {$in}^3$
The weight of the jar is simply its volume times its density. We have to convert density of the glass:
$165 \frac{lb}{ft^3}=\frac{165}{1728}\frac{lb}{in^3}=0.0949\frac{lb}{in^3}$
Now we can find the mass:
$m_j=V_s\rho_j=27.17\cdot 0.0949=2.58 $lb$
The volume of water jar would displace depends how far the jar would sink.
There two forces acting on the jar when you put in the water. The first one is gravity and it's pulling the jar down. The second force is the buoyancy the and this force pushes the jar upward. In order for that jar to be stable, these two forces must be the same.
The buoyancy is given with this formula:
$B=\rho gV_{disp}$
The volume of water that would be displaced by the jar can be expressed by this formula:
$V_{disp}=\pi h'(r+a)^2$
Where h' is the height of the jar that is submerged in the water.
So the buoyancy force is:
$B=\rho_w g\pi h'(r+a)^2$
This force must be equal to the force of gravity acting on the jar:
$F_g=B\\ m_jg=\rho_w g\pi h'(r+a)^2\\ m_j=\rho_w \pi h'(r+a)^2\\$
We solve for h':
$h'=\frac{m_j}{\rho_w \pi(r+a)^2}$
If we plug in the number we get:
$h'=2.24 $in$
Now we can calcuate the amount of water discplaced and its mas(keep in mind that we also converted density of water into lb/in^3):
$V_{disp}=\pi h'(r+a)^2=71.50${in}^3\\
m_v=\rho_w\cdot V_{disp}=2.58$lb$
We get the same mass as the mass of the jar, which makes perfect sense:
$m_jg=\rho_w V_{disp}g\\
m_j=m_{wdisp}$

4 0

4 years ago